Large nuclei are unstable for a number of reasons one being that the long range coulomb repulsion force between protons is becoming more "dominant", as the number of protons in the nucleus increases, compared with the attractive short range nuclear force between all the nucleons.
So large nuclei try and dilute the effect of the coulomb repulsion by adding more neutral neutrons which feel the attractive strong nuclear force but not the coulomb force.
But it is a delicate balance and large nuclei want to become more stable by turning into nuclei with smaller numbers of protons within them.
The parameter which is useful when comparing the stability of nuclei is the binding energy per nucleon and the largest values of this parameter centre around the nucleus of iron.
So nuclei much larger than iron are trying to break up into smaller nuclei which are closer in composition to iron nuclei and thus become more stable.
In the fission process a large unstable nucleus splits up into two more stable nuclei and some neutrons and in the process energy is released and manifests itself as kinetic energy of the fission products
The masses of the nucleus undergoing fusion is greater than the mass of the fission products and that mass difference is called the mass defect $\Delta m$ and the energy released in the fission process is given by $\Delta m \, c^2$.
So your statement
If the mass defect turned into binding energy
is not correct and the energy equivalent of the mass defect is released in the fission process but indeed the resulting nuclei are more stable (have a higher binding energy per nucleon) than the nucleus undergoing fission.
Update as a result of a question frpm @BøbbyLeung
If the mass defect does not become binding energy, then where would the increase in binding energy of the product nuclei come from (based on the conservation of energy)?
The binding energy of a nucleus is the energy required to break up a nucleus into its constituent parts (individual nucleons).
In general a smaller nucleus has a smaller binding energy than a larger nucleus because there are more interactions ("bonds") between the nucleons but the "bonds" between the nucleons in the nuclei around iron are stronger - it requires more energy to remove a nucleon from a small nucleus than a large nucleus.
In terms of numbers let us suppose that you had a nucleus with $120$ nucleons and the binding energy per nucleon was $6$ arbitrary energy units .
If this nucleus was assembled from individual nucleons $120 \times 6 = 720$ units of energy would have been released.
Suppose further that for a nucleus with $240$ nucleons the binding energy was $5$ energy units.
To assemble such a nucleus from individual nucleons would have released $240 \times 5 = 1200$ units of energy.
Now what happens when the nucleus with $240$ nucleons splits into (undergoes fission) two nuclei each with $120$ nucleons.
The binding energy started at $1200$ and ends up as $2 \times 720 = 1440$.
This is an increase in the binding energy ie it takes more energy to break up two $120$ nucleon nuclei than one $240$ nucleon nucleus.
Conservation of energy requires that the fission of a $240$ nucleon nucleus into two $120$ nucleon nuclei must result in the release of $1440 - 1200 = 240$ units of energy.
but may I ask if the mass defect isnt transformed into binding energy, then why does the binding energy of the fission product increase (are we getting free binding energy for nothing)? And why can we calculate the energy released in a nuclear reaction by minusing the binding energies of the products by the binding energies of the reactant?
Perhaps your questions are answered in terms of an energy level diagram?